Divide using synthetic division $$ ( 2x^{5}-6x^{4}+2x^{3}+4x^{2}-5x+3 ) \div ( 0 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}0&2&-6&2&4&-5&3\\& & 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{4}&\color{blue}{-5}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-6x^{4}+2x^{3}+4x^{2}-5x+3 }{ x } = \color{blue}{2x^{4}-6x^{3}+2x^{2}+4x-5} ~+~ \frac{ \color{red}{ 3 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}0&\color{orangered}{ 2 }&-6&2&4&-5&3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & \color{blue}{0} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}0&2&\color{orangered}{ -6 }&2&4&-5&3\\& & \color{orangered}{0} & & & & \\ \hline &2&\color{orangered}{-6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & 0& \color{blue}{0} & & & \\ \hline &2&\color{blue}{-6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}0&2&-6&\color{orangered}{ 2 }&4&-5&3\\& & 0& \color{orangered}{0} & & & \\ \hline &2&-6&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & 0& 0& \color{blue}{0} & & \\ \hline &2&-6&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}0&2&-6&2&\color{orangered}{ 4 }&-5&3\\& & 0& 0& \color{orangered}{0} & & \\ \hline &2&-6&2&\color{orangered}{4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & 0& 0& 0& \color{blue}{0} & \\ \hline &2&-6&2&\color{blue}{4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}0&2&-6&2&4&\color{orangered}{ -5 }&3\\& & 0& 0& 0& \color{orangered}{0} & \\ \hline &2&-6&2&4&\color{orangered}{-5}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&2&-6&2&4&-5&3\\& & 0& 0& 0& 0& \color{blue}{0} \\ \hline &2&-6&2&4&\color{blue}{-5}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}0&2&-6&2&4&-5&\color{orangered}{ 3 }\\& & 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{4}&\color{blue}{-5}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-6x^{3}+2x^{2}+4x-5 } $ with a remainder of $ \color{red}{ 3 } $.